Let $a$, $b$, and $c$ be the $3$ roots of $x^3-x+1=0$.  Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
Solution: We can substitute $x=y-1$ to obtain a polynomial having roots $a+1$, $b+1$, $c+1$, namely,
\[(y-1)^3-(y-1)+1=y^3-3y^2+2y+1.\]The sum of the reciprocals of the roots of this polynomial is, by Vieta's formulas, $\frac{2}{-1}=\boxed{-2}$.